vec{a}, vec{b}, vec{c} are non-coplanar vecto | vedclass

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(A) Given the line $\vec{r}=\vec{a}+2 \vec{b}+p(\vec{a}-2 \vec{c}) \quad \dots(1)$ and the plane $\vec{r}=3 \vec{a}-q(\vec{c}-\vec{b})+k(\vec{a}-\vec{

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$-8$

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(A) Given the line $\vec{r}=\vec{a}+2 \vec{b}+p(\vec{a}-2 \vec{c}) \quad \dots(1)$ and the plane $\vec{r}=3 \vec{a}-q(\vec{c}-\vec{b})+k(\vec{a}-\vec{b}+\vec{c}) \quad \dots(2)$.Equating the two expressions for $\vec{r}$:$\vec{a}+2 \vec{b}+p \vec{a}-2p \vec{c} = 3 \vec{a}-q \vec{c}+q \vec{b}+k \vec{a}-k \vec{b}+k \vec{c}$$\vec{a}(1+p) + 2 \vec{b} - 2p \vec{c} = \vec{a}(3+k) + \vec{b}(q-k) + \vec{c}(k-q)$Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,we compare the coefficients:$1+p = 3+k \Rightarrow p-k = 2$$2 = q-k \Rightarrow q-k = 2$$-2p = k-q \Rightarrow q-k = 2p$From $q-k=2$ and $q-k=2p$,we get $2p=2 \Rightarrow p=1$.Substituting $p=1$ into $p-k=2$,we get $1-k=2 \Rightarrow k=-1$.Substituting $k=-1$ into $q-k=2$,we get $q-(-1)=2 \Rightarrow q=1$.Now,substitute $p=1$ into equation $(1)$ to find the position vector $\vec{r}$:$\vec{r} = \vec{a}+2 \vec{b}+1(\vec{a}-2 \vec{c}) = 2 \vec{a}+2 \vec{b}-2 \vec{c}$.Comparing this with $\vec{r}=x \vec{a}+y \vec{b}+z \vec{c}$,we get $x=2, y=2, z=-2$.Therefore,$x y z = 2 	imes 2 	imes (-2) = -8$.

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